# NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

NCERT solutions for class 8 maths chapter 8 Comparing Quantities- This chapter is a combination of several important topics, like ratios, percentages, discount, profit & loss and simple and compound interest. To help the students, step by step solutions of NCERT class 8 maths chapter 8 comparing quantities are provided. The ratio of two quantities is known as comparing quantity. In this chapter, we are going to compare quantities like marks of two students, the height of two persons, a comparison of profit, sales value, etc. In the CBSE NCERT solutions for class 8 maths chapter 8 comparing quantities, there are 3 exercises with 26 questions. The first exercise of CBSE class 8 chapter comparing quantities is on percentage and ratio.

The second exercise is on finding the increase or decrease percent and finding discounts. Through these NCERT solutions for class 8 maths chapter 8 comparing quantities, students can cover all the problems related to ratios, percentages, discount, profit-loss, and simple-compound interest. The solutions of NCERT help students to practice questions on the concept studied in a chapter.

## NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.1

### In a primary school, the parents were asked the number of hours they spend per day in helping their children to do homework. There were 90 parents who helped $\frac{1}{2}$ hour to $1\frac{1}{2}$hours. The distribution of parents according to the time for which they said they helped is given in the adjoining figure; 20% helped for more than $1\frac{1}{2}$  hours; 30% helped for   $\frac{1}{2}$  hour to $1\frac{1}{2}$   hours; 50% did not help at all.

Using this, answer the following :

(i) How many parents were surveyed?

Let the number of parents surveyed be X.

30% of them helped for   $\frac{1}{2}$  hour to $1\frac{1}{2}$   hours.

In total, there were 90 such parents.

$\therefore$ 30% of X = 90

$\Rightarrow \frac{30}{100}\times X = 90 \\ \Rightarrow X = 90\times100 /30 = 300$

Therefore, there were a total of 300 parents who were surveyed.

CBSE NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.3

(a) A dress marked at Rs 120.

The marked price of dress = Rs120

Discount rate = 20%

$\\\\ \\Total\ discount =Marked\ Price\times \frac{Discount\ rate}{100}\\ \\Total\ discount=120\times \frac{20}{100}\\ \\Total\ discount=Rs.24\\$

Selling price (SP) = Marked price - Total discount = Rs (120- 24) = Rs 96.

## 1(b)  A shop gives 20% discount. What would the sale price of each of these be?

(b) A pair of shoes marked at Rs 750

The marked price of shoes=Rs 750

Discount rate = 20%

$\\Total\ discount =Marked\ Price\times \frac{Discount\ rate}{100}\\ \\Total\ discount=750\times \frac{20}{100}\\ \\Total\ discount=Rs.150\\$

Selling price= Marked price- Total discount =Rs (750- 150) = Rs 600.

(c)  A bag marked at Rs 250

The marked price of bag = Rs 250

Discount rate=20%

$\\Total\ discount =Marked\ Price\times \frac{Discount\ rate}{100}\\ \\Total\ discount=250\times \frac{20}{100}\\ \\Total\ discount=Rs.50\\$

Selling price = Marked price - Total discount = Rs (250- 50) = Rs 200.

Marked price = Rs 15000

Selling Price = Rs 14400

We know, Discount amount = Marked price - Selling price

Discount amount = Rs 15000-Rs 14400 = Rs 600

$\\Discount\ percent=\frac{Discount}{Marked\ Price}\times 100\\ \therefore Discount\ percent=\frac{600}{15000}\times 100=4\%\\$

Therefore, Discount percentage = 4%

Selling price = Rs 5225

Discount percent = 5%

$\\ Selling\: price = Marked \:price - Discount \:Amount\\ Selling\: price=Marked\ price-Marked\ price\times \frac{5}{100}\\ Selling\: price=Marked price\times \frac{95}{100}\\ \\Marked\ price= Selling\ price\times \frac{100}{95}\\ \\Marked\ price=\frac{100}{95} \times 5225\\$

Therefore, the marked price = Rs 5500

Solutions of NCERT class 8 maths chapter 8 comparing quantities topic 8.4.1

(a) A cycle of Rs 700 with Rs 50 as overhead charges.

$\therefore$ Cost price = Rs 700 + Rs 50 = Rs 750

Profit percent = 5%

We know,

$\dpi{100} \\Profit= Cost\ price\times \frac{Profit\ percent}{100} \\Profit=750\times \frac{5}{100}$

$\therefore$ Profit = Rs 37.5

Now, Selling price (SP) = Cost price + Profit =Rs 750 + Rs 37.5 = Rs 787.50

Therefore, the selling price (SP) is Rs 787.50

(b) a lawnmower bought at Rs 1150 with Rs 50 as transportation charges.

If a profit of 5% is made on a lawnmower bought at Rs 1150 with Rs 50 as transportation charges then selling price is:

$\\Cost\ price=Rs\ 1150+Rs\ 50\\ \\Cost\ price=Rs\ 1200\\ \\Profit\ percent=5 \\Profit= Cost\ price\times \frac{Profit\ percent}{100} \\Profit=1200\times \frac{5}{100} \\Profit=Rs\ 60 \\Selling\ price= Cost\ price+Profit \\Selling\ price=Rs\ 1200+Rs\ 60 \\Selling\ price=Rs\ 1260$

(c) a fan bought at Rs 560 and expenses of Rs 40 made on its repairs.

$\therefore$ Cost price = Rs 560 + Rs 40 = Rs 600

Profit percent = 5%

$\dpi{100} \\Profit= Cost\ price\times \frac{Profit\ percent}{100} \\Profit=600\times \frac{5}{100}$

$\therefore$ Profit = Rs 30

Now, Selling price (SP) = Cost price + Profit =Rs 600 + Rs 30 = Rs 630

Therefore, the selling price (SP) is Rs 630

$\\Profit\ on\ first\ TV=10000\times \frac{10}{100}=Rs\ 1000\\ \\Loss\ on\ second\ TV=10000\times \frac{10}{100}=Rs\ 1000\\$

Net profit = Rs 1000 - Rs 1000 = 0

He neither made an overall profit or loss since the profit made on the first tv equals the loss suffered on the second one.

Let the number be x.

Half the number = $\frac{x}{2}$

$\\Decrease\ in\ percent=\frac{Number-Half\ the\ number}{Number}\times 100\\ =\frac{x-\frac{x}{2}}{x}\times 100\\ =\frac{1}{2}\times 100\\$

= 50 %

Therefore, If we take half the number, the decrease in percent is 50%

Difference between Rs 2400 and Rs 2000 = Rs 2400- Rs 2000 = Rs 400

$\therefore$  The per cent by which Rs 2,000 is less than Rs 2,400 is

$\\ \frac{400}{2400}\times 100=16.66 \%\\$ (Less with respect to 2400. Hence, 2400 will be in the denominator!)

Rs 2000 is less than Rs 2400 by 16.66%

Now,

$\therefore$  The per cent by which Rs 2,400 is more than Rs 2,000 is

$\frac{400}{2000}\times 100=20 \%$        (More with respect to 2000. Hence, 2000 will be in the denominator!)

Rs 2400 is more than Rs 2000 by 20%

Therefore, they are not the same.

## Q Find interest and amount to be paid on Rs 15000 at 5% per annum after 2 years.

$\\Interest\ =Principal\ amount\times\frac{Interest\ rate}{100}\times time\\ \:\:=15000\times \frac{5}{100}\times 2\\$

=Rs1500

Amount to be paid = Interest + Principal Amount

=Rs 15000 + Rs 1500

=Rs 16500

Find the time period and rate for each.

1. A sum taken for  $1\frac{1}{2}$ years at 8% per annum is compounded half yearly.

Since the sum taken is compounded half yearly:

$\\Since\ the\ sum\ taken\ is\ compounded\ half\ yearly:\\ Time\ period=2\times 1\frac{1}{2}\\ Time\ period=3\\ Rate=\frac{8}{2}=4\%$

Time period = 3 half years

Rate = 4% per half year

## Find the time period and rate for each.

A sum taken for 2 years at 4% per annum compounded half yearly.

Since the sum taken is compounded half yearly:

$\\Since\ the\ sum\ taken\ is\ compounded\ half\ yearly:\\ Time\ period=2\times 2\\ Time\ period=4\\ Rate=\frac{4}{2}=2\%$

Time period = 4 half years.

Rate = 2% per half year

Find the amount to be paid

$\\A_{n}=P_{1}(1+\frac{R}{100})^{n}\\ P_{1}=Rs\ 2400\\ n=2\\ R=5\\ A_{2}=2400\times (1+\frac{5}{100})^2\\ A_{2}=Rs\ 2646\\$

The amount to be paid at the end of 2 years is Rs 2646

Find the amount to be paid

At the end of 1 year on Rs. 1800 at 8% per annum compounded quarterly.

$\\A_{n}=P_{1}(1+\frac{R}{100})^{n}\\ P_{1}=Rs\ 1800\\ n=4\\ R=\frac{8}{4}=2\ percent\ per\ quarter\ year\\ A_{4}=1800\times (1+\frac{2}{100})^4\\ A_{4}=Rs\ 1948.37\\$

The Amount to be paid at the end of 4 years is Rs 1948.37

Solutions of NCERT class 8 maths chapter 8 comparing quantities topic 8.9

Price of machinery = Rs. 10,500

$Depriciation=10500\times \frac{5}{100}=Rs.\ 525$

Value after one year =  Rs 10,500- Rs 525 = Rs  9975

The current population of city

= 12 lakh

1200000

Population after two years =

$P\left ( 1+\frac{R}{100} \right )^{n}$

$= 1200000\left ( 1+\frac{4}{100} \right )^{2}$

$= 1200000\left ( 1+0.04 \right )^{2}$

$= 1200000\left ( 1.04 \right )^{2}$

$= 1200000\times 1.0816$

$= 1297920$

Thus, the population after two years is 1297920.

NCERT solutions for class 8 maths chapter 8 comparing quantities exercise 8.1

Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

The ratio of the speed of cycle to the speed of scooter =

$\frac{15\ km/ hour}{30\ km/ hour} = \frac{15}{30} = \frac{1}{2}$

= 1:2

5m to 10km

To find the ratio, we need to make both quantities of the same unit.

We know, 1 km = 1000 m.

$\therefore$ 10 km = 10x1000 m = 10000 m

Therefore the required ratio =

$\frac{5m}{10\times1000m} = \frac{5}{10000} = \frac{1}{2000}$

= 1 : 2000

50 paise to Rs. 5

To find the ratio, we first make the quantities of the same unit.

We know, Rs. 1 = 100 paisa. $\therefore$ Rs. 5 = 5x100 = 500 paisa

Therefore, the required ratio =

$\frac{50 paisa}{Rs. 5}= \frac{50 paisa}{5\times100 paisa} = \frac{1}{10}$

= 1:10

(a) $3:4$

To convert a ratio to a percentage, we multiply the ratio by 100%.

$\therefore$ The required percentage of the ratio 3:4 =

$\frac{3}{4}\times100\ \%= 75\%$

(b) $2:3$

To convert a ratio to a percentage, we multiply the ratio by 100%.

$\therefore$ The required percentage of ratio 2:3 =

$\frac{2}{3}\times100\ \%= 66.67 \%$

Given;

72% of 25 students are interested in mathematics

$\therefore$ Percentage of students not interested in mathematics = (100 - 72)% = 28 %

$\therefore$ The number of students not interested in mathematics = 28% of 25

$=\frac{28}{100}\times25 = \frac{28}{4}$

$= 7$

Therefore, 7 students (out of 25) are not interested in mathematics.

Given,

Win percentage of the team = 40%

This means that they won 40 matches out of 100 matches played.

$\therefore$ They won 10 matches out of  $\frac{100}{40}\times10$ = 25 matches played.

Therefore, they played a total of 25 matches in all.

Let the amount of money Chameli had in the beginning = Rs. X

She spends 75% of the money.

$\therefore$ Percentage of money left = (100 - 75)% = 25%

Since she has Rs. 600 left.

$\therefore$ 25% of X = 600

$\Rightarrow \frac{25}{100}\times X = 600 \\$

$\Rightarrow \frac{X}{4} = 600$

$\\ \Rightarrow X = 600\times4$

X =2400

Therefore, Chameli had Rs. 2400 with her in the beginning.

Given,

Total number of people = 50 lakhs

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

Since remaining people like other games,

$\therefore$ Percentage of people who like other games = {100 - (60 + 30) } = (100 - 90) = 10%

10% of the people like other games.

Now,

$\therefore$ Number of people who like cricket = 60% of 50 lakhs

$=\frac{60}{100}\times50= 30 lakhs$

$\therefore$ Number of people who like football = 30% of 50 lakhs

$=\frac{30}{100}\times50= 15 lakhs$

$\therefore$ Number of people who like other games = 10% of 50 lakhs

$=\frac{10}{100}\times50 = 5 lakhs$

## CBSE NCERT solutions for class 8 maths chapter 8 comparing quantities exercise 8.2

Given,

Percentage increase in the salary = 10%

Therefore, if the original salary was Rs. 100, the new salary is Rs. 110

If new salary is Rs 1,54,000, the original salary was

$=Rs. \frac{100}{110}\times154000= Rs. 140000$

Therefore, the original salary was Rs. 14,00,00.

Given,

Number of people who went to the zoo on Sunday = 845

Number of people who went to the zoo on Monday = 169

$\therefore$ The decrease in the number of people visiting the zoo = (845 - 169) = 676

$\therefore$ Percentage decrease

$=\frac{676}{845}\times100 \%$

$=80\%$    (Decrease from the original number)

The percent decrease in the people visiting the Zoo is 80%

Given,

Cost price (CP) of the 80 articles = Rs. 2,400

Profit percentage = 16%

$\therefore$ Profit amount on all 80 articles = 16% of 2400 = 16/100 x 2400 = Rs. 384

$\therefore$ The selling price of the 80 articles = Rs. (2400 + 384) = Rs. 2784

$\therefore$ Selling price (SP) of each item = Rs. 2784/80 = Rs. 34.8

Therefore, the selling price of one article is Rs.34.8

## 4  The cost of an article was Rs 15,500. Rs 450 was spent on its repairs. If it is sold on a profit of 15%, find the selling price of one article.

Given,

Cost of the article = Rs. 15500

Cost of repair = Rs. 450

$\therefore$ Cost price(CP) of the article = Rs. (15500 + 450) = Rs. 15950

Profit percentage = 15%

$\therefore$ Profit amount = 15% of Rs.15950 = Rs. 2392.50

$\therefore$ Selling Price = Rs.(15950 + 2392.50) = Rs. 18342.50

OR

$\therefore$ Selling price (SP) of the article = CP + Profit = CP + (15% of CP)

= 115% of CP

$=\frac{115}{100}\times15950= Rs. 18342.50$

Given,

The cost price of the VCR = Rs. 8000

The cost price of TV = Rs. 8000

Now, He made a loss of 4% on VCR.

$\therefore$ Selling price (SP) of the VCR = (100-4)% of CP = 96% x Rs. 8000 = Rs. 7680

Again, He made a profit of 8% on TV.

$\therefore$ Selling price (SP) of the TV = (100+8)% of CP = 108% x Rs. 8000 = Rs. 8640

Net Selling price = Rs. (7680 + 8640) = Rs. 16320

Net Cost price = Rs. (2x8000) = Rs. 16000

Since, SP > CP, he made a net profit (gain)

Now, Net Gain = CP - SP = Rs. (16320-16000) = Rs. 320

$\therefore$ Gain % = $\frac{Profit}{CP}\times100\% = \frac{320}{16000}\times100\%$ = 2%

$\therefore$ The shopkeeper made a gain of 2% on the whole transaction.

## 6 During a sale, a shop offered a discount of 10% on the market prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs 1450 and two shirts marked at Rs 850 each?

Since the 10% discount is on all the items, we can calculate the Selling price by totalling the Cost price of all item bought.

Now,

Total Cost price(CP) of the items he bought = CP of a pair of jeans + CP of two shirts = Rs.(1450 + 850 + 850) = Rs. 3150

$\therefore$ The selling price of these items = (100- 10)% of Rs. 3150 = 90% x Rs.3150 = Rs. 2835

$\therefore$ The customer has to pay Rs. 2835.

Given,

The milkman sold two of his buffaloes for Rs 20,000 each. This is the Selling price of the buffaloes.

Let CP of one of the buffalo be Rs. X and the other be Rs. Y

Since he made a profit of 5% on one of them.

$\therefore$ 105% of X = Rs. 20000

$\\ \implies \frac{105}{100}\times X = 20000 \\ \implies X = \frac{100}{105}\times20000$

$\therefore$ X = Rs. 19047.6

Similarly, since he made a loss of 10% on the other.

$\therefore$ 90% of Y = Rs. 20000

$\\ \implies \frac{90}{100}\times Y = 20000 \\ \implies Y = \frac{100}{90}\times20000$

$\therefore$ Y = Rs. 22222.2

$\therefore$ Net CP = Rs.(19047.6 + 22222.2) = Rs. 41269.8

And net SP = 2 x Rs.20000 = Rs. 40000

Since SP < CP, he made a net loss.

$\therefore$ His overall loss = Rs. (41269.8 - 40000) = Rs. 1269.8 = Rs. 1270 (approx)

Given,

Cost price of the TV = Rs. 13000

Sales tax at the rate of 12%

$\therefore$ Selling price = CP + Sales Tax = CP + 12% of CP

= 112% of CP =

$\frac{112}{100}\times 13000 = Rs. 14560$

$\therefore$ Vinod will have to pay Rs. 14,560.

The discount given was 20% which means if CP is Rs. 100 then the SP is Rs. 80

$\therefore$ If SP is Rs. 80 then CP is Rs. 100

For SP of Rs. 1600, CP

$=\frac{100}{80}\times1600 = Rs. 2000$

$\therefore$ The marked price is Rs. 2000.

Given,

VAT = 8%

Let the original price be Rs. 100

Original price +  VAT  = Rs. 100 + Rs. $\left ( 8\%\, of\, 100 \right )$

Original price +  VAT  = Rs. 100 + Rs. 8 = Rs. 108

$\therefore$ If the price after VAT is Rs.5400, then the price before VAT is  $Rs. \frac{100}{108}\times5400 = Rs. 5000$

$\therefore$ The price before VAT was added is Rs. 5000.

Given,

GST = 18%

Cost with GST included = Rs. 1239

Cost without GST  = x Rs.

$x + ( \frac{18}{100} \times x ) = 1239$

cost before GST+ GST = cost with GST

$x + ( \frac{9x}{50} ) = 1239$

x = 1050

Price before GST = 1050 rupees

## Solutions of NCERT class 8 maths chapter 8 comparing quantities exercise 8.3

Given,

Principal,P = Rs 10800

Compound Interese Rate,R =   $12\frac{1}{2} \% = \frac{25}{2}\%$ p.a.

Time period,n = 3 years.

We know,

Amount when interest is compounded annually, A = $P(1+\frac{R}{100})^n$

Therefore, the required amount = $10800(1+\frac{25}{2\times100})^{3}$  = Rs. 15377.34

And Compound Interest, CI = Amount - Principal = Rs. (15377.34 - 10800) = Rs. 4577.34

## 1 (b)  Calculate the amount and compound interest on

Given,

Principal,P =Rs.18000,  Rate,R = 10%   and time period,n = 2.5 years.

We know, Amount when interest is compounded annually = $P(1+\frac{R}{100})^n$

Amount after 2 years at 10% , A = $18000(1+\frac{10}{100})^2$ = Rs.21780

This acts as the principal amount for the next half year.

SI on next 1/2 year at = $\frac{21780\times\frac{1}{2}\times10}{100}$= Rs. 1089

Therefore, Total amount to be paid after 2.5 years = Rs. (21780+1089) = Rs.22869

Now, Compound Interest after 2 years = A - P = Rs.(21780-18000) = Rs. 3780

Therefore, Compound Interest after 2.5 years, CI = Rs. 3780 + SI = Rs.4869

Given,

Principal,P =Rs 62500,

Compound interest Rate,R = 8% compounded half yearly for 1.5 years.

Since it is compounded half yearly, R becomes half = 4%, and time period doubles, n = 3 years.

We know, Amount when interest is compounded annually, A = $P(1+\frac{R}{100})^n$

Therefore, the required amount = $62500(1+\frac{4}{100})^3$  = Rs.70304

And Compound Interest, CI = Amount - Principal = Rs. (70304 - 62500) = Rs. 7804

(d) Rs 8000 for 1 year at 9% per annum compounded half yearly.   (You could use the year by year calculation using SI formula to verify)

Given,

Principal,P =Rs.8000,  Rate, R =   9% per annum compounded half yearly for 1 year.

Now, there are two half years in a year. Therefore compounding has to be 2 times.

And rate = half of 9% = 4.5% half yearly.

Therefore, the required amount = $8000(1+\frac{9}{2\times100})^{2}$  = Rs. 8736.20

And Compound Interest, CI = Amount - Principal = Rs. (8736.20 - 8000) = Rs. 736.20

Given,

Principal, P =Rs.10000,  Rate, R = 8% per annum compounded half yearly for 1 year.

Now, there are two half years in a year. Therefore compounding has to be 2 times.

And rate = half of 10% = 5% half yearly.

Therefore, the required amount = $10000(1+\frac{4}{100})^{2}$  = Rs. 10816

And Compound Interest, C.I. = Amount - Principal = Rs. (10816 - 10000) = Rs. 816.

The amount borrowed from the bank = Principal amount, P = Rs 26400

Compound interest rate, R = 15% p.a.

Time period = 2 years 4 months = $2\frac{4}{12} = 2\frac{1}{3} years$

We know, Amount when interest is compounded annually, A = $P(1+\frac{R}{100})^n$

Therefore, for the first 2 years, amount, A = $26400(1+\frac{15}{100})^2$ = Rs 34914

Now, this would act as principal for the next 1/3 year. We find the SI on Rs 34914 for 1/3 year.

SI = $\frac{34914\times\frac{1}{3}\times15}{100}$ = Rs 1745.70

Therefore, Required amount at the end of 2 years and 4 months = A + SI = Rs (34914 + 1745.70) = Rs 36659.70

For Fabina,

Principal,P =Rs 12500

Simple interest Rate,R = 12% p.a.

Time period,n = 3 years.

$\therefore$ Simple Interest, SI at 12% for 3 years = $3\times\frac{12500\times12}{100}$ = Rs 4500

Principal,P =Rs 12500

Compound interest Rate,R = 10% p.a.

Time period,n = 3 years.

We know, Amount when interest is compounded annually,

$A =P(1+\frac{R}{100})^n$

$A=12500(1+\frac{10}{100})^3 = Rs 16637.50$

$\therefore Compound \:\:Interest, CI = A - P = Rs\: (16637.50 - 12500) = Rs \:4137.50$

$\therefore$ Fabina pays more interest and Rs (4500 - 4137.50) = Rs 362.50 more.

## 4. I borrowed Rs 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Given,

Principal,P =Rs 12000

Simple interest Rate,R = 6% p.a.

Time period,n = 2 years.

$\therefore$ Simple Interest, SI at 6% for 2 years =

$=2\times\frac{12000\times6}{100} = Rs 1440$

If he would have borrowed it at a compound interest rate, R = 6% p.a.

We know, Amount when interest is compounded annually, A =

$A = P(1+\frac{R}{100})^n$

$\therefore A = 12000(1+\frac{6}{100})^2= Rs 13483.20$

$\therefore Compound\:\: Interest, CI = A - P = Rs \:(13483.20 - 12000) = Rs\: 1483.20$

$\therefore$ He would have to pay Rs (1483.20 - 1440) = Rs 43.20 extra.

(i) after 6 months?

Given,

Principal, P = Rs 60,000

Compound interest rate, R = 12% p.a

= 6 % half yearly

For a period of 6 months. $\therefore$ Time period, n = 1 half year (As there is 1 half year in 6 months.)

We know, Amount when interest is compounded annually, A =

$A = P(1+\frac{R}{100})^n$

$A =60000(1+\frac{6}{100})^1= Rs 63600$

After 6 months, Vasudevan would get an amount Rs. 63600.

(ii) after 1 year?

Given,

Principal, P = Rs 60,000

Compound interest rate, R = 12% p.a

= 6 % half-yearly

For a period of 1 year. $\therefore$ Time period, n = 2 half years (As there are 2 half years in a year.)

We know, Amount when interest is compounded annually, A =

$A = P(1+\frac{R}{100})^n$

$A = 60000(1+\frac{6}{100})^2= Rs 67416$

After 1 year, Vasudevan would get an amount Rs. 67416.

(i) compounded annually.

(ii) compounded half yearly.

(i) Given,

Principal amount, P = Rs 80000

Rate of interest, R = 10% p.a.

Time period = $1\frac{1}{2}$ years.

We know, Amount when interest is compounded annually, A =

$A =P(1+\frac{R}{100})^n$

Now, For the first year, A=

$A =80000(1+\frac{10}{100})^1= Rs. 88000$

For the next half-year, this will act as the principal amount.

$\therefore$ Interest for 1/2 year at 10% p.a =

$=\frac{88000\times\frac{1}{2}\times10}{100}= Rs 4400$

Required total amount = Rs (88000 + 4400) = Rs 92400

(ii) If it is compounded half-yearly, then there are 3 half years in $1\frac{1}{2}$ years.

$\therefore$ n = 3 half years.

And, Rate of interest = half of 10% p.a = 5% half yearly

$\therefore A =80000(1+\frac{5}{100})^3= Rs.\: 92610$

$\therefore$ The difference in the two amounts = Rs (92610 - 92400) = Rs 210

Given,

Principal amount, P = Rs 8,000

Compound rate of interest, R = 5% p.a.

Time period, n = 2 years

We know, Amount when interest is compounded annually,

$A =P(1+\frac{R}{100})^n$

$A =8000(1+\frac{5}{100})^2= Rs.\: 8,820$

Therefore, the amount credited against her name at the end of the second year is Rs 8,820

(ii) The interest for the 3rd year.

Now, the amount after 2nd year will become the principal amount for the 3rd year

$\therefore$ Principal amount, P = Rs 8,820

Compound rate of interest, R = 5% p.a.

Time period, n' = 1 year

$\therefore Interest \:for\: the \:3rd \:year =\frac{8820\times1\times5}{100} = Rs.\ 441$

Therefore, the interest for the 3rd year is Rs 441.

Principal = Rs.10,000

Time =   $1\frac{1}{2}$   years

Rate = 10% per annum

CASE 1  Interest on compounded half yearly.

Rate = 10% per annum  = 5 % per half yearly

$T= 1\frac{1}{2} years,n=3$

$A= P\left ( 1+\frac{R}{100} \right )^{n}$

$A= 10000\left ( 1+\frac{5}{100} \right )^{3}$

$A= 10000\left ( 1+0.05 \right )^{3}$

$A= 10000\left ( 1.05 \right )^{3}$

$A= 11576.25$ = Amount

CI = Amount - principal

CI  = $11576.25-10000$

CI = 1576.25

CASE 2     Interest on compounded anually

Rate = 10% per annum

$T= 1\frac{1}{2} years ,n=1$

$A= P\left ( 1+\frac{R}{100} \right )^{n}$

$A= 10000\left ( 1+\frac{10}{100} \right )^{1}$

$A= 10000\left ( 1+0.1\right )^{1}$

$A= 10000\left ( 1.1\right )^{1}$

$A= 11000$ = Amount

CI = Amount - principal

CI  = $11000-10000$

CI = 1000

Interest for half years on 11000 =

$\frac{P\times R\times T}{100}$

$=\frac{11000\times 10\times \frac{1}{2}}{100}$

$= 55\times 10$

= 550

Total interest = $1000+550$

= RS 1550

Since 1576.25 $>$ 1000

Thus, interest would be more in CASE 1 i.e. compounded half yearly

Given,

Principal amount, P = Rs 4,096

Rate of interest, R

$R =12\frac{1}{2} \% = \frac{25}{2} \% p.a. =\frac{25}{4} \% \ half\ yearly$

Time period, n = 18 months = (12 + 6) months = 1.5 years = 3 half years

(There are 3 half years in 1.5 years)

We know,

Amount when interest is compounded annually, (A)

$A =P(1+\frac{R}{100})^n$

Therefore, the required amount

$=4096(1+\frac{25}{4\times100})^{3} = 4913$

$\therefore$ Ram will get Rs 4,913 after 18 months.

## 10 The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) find the population in 2001.

Let the population in 2001 be P

Compound rate of increase = 5% p.a.

The population in 2003 will be more than in 2001

Time period, n = 2 years    (2001 to 2003)

$\therefore$ $\\ 54000 = P(1 + \frac{5}{100})^2 \\ \implies P = \frac{54000\times100\times100}{105\times105} \approx 48980$

Therefore, the population in 2001 was 48980 (approx)

(ii) what would be its population in 2005?

Let the population in 2001 be P'

Compound rate of increase = 5% p.a.

The population in 2005 will be more than in 2003

Time period, n = 2 years    (2003 to 2005)

$\therefore$ $\\ P' = 54000(1 + \frac{5}{100})^2 \\ \implies P' = \frac{54000\times105\times105}{100\times100} \approx 59535$

Therefore, the population in 2005 will be 59535 (approx)

Given,

Initial count of bacteria, P = 5, 06,000 (Principal Amount)

Rate of increase, R = 2.5% per hour.

Time period, n = 2 hours

(This question is done in a similar manner as compound interest)

Number of bacteria after 2 hours = $P(1+\frac{R}{100})^n$

$= 506000(1+\frac{2.5}{100})^2 \approx 531616$

Therefore, the number of bacteria at the end of 2 hours will be 531616 (approx)

Given,

Principal = Rs 42,000

Rate of depreciation = 8% p.a

$\therefore$ Reduction = 8% of Rs 42000 per year

$=\frac{42000\times 8\times 1}{100} = Rs\ 3360$

$\therefore$ Value at the end of 1 year = Rs (42000 – 3360) = Rs 38,640

## NCERT solutions for class 8 maths chapter wise

 Chapter -1 NCERT solutions for class 8 maths chapter 1 Rational Numbers Chapter -2 Solutions of NCERT for class 8 maths chapter 2 Linear Equations in One Variable Chapter-3 CBSE NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Chapter-4 NCERT solutions for class 8 maths chapter 4 Practical Geometry Chapter-5 Solutions of NCERT for class 8 maths chapter 5 Data Handling Chapter-6 CBSE NCERT solutions for class 8 maths chapter 6 Squares and Square Roots Chapter-7 NCERT solutions for class 8 maths chapter 7 Cubes and Cube Roots Chapter-8 NCERT solutions for class 8 maths chapter 8 Comparing Quantities Chapter-9 NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities Chapter-10 CBSE NCERT solutions for class 8 maths chapter 10 Visualizing Solid Shapes Chapter-11 NCERT solutions for class 8 maths chapter 11 Mensuration Chapter-12 NCERT Solutions for class 8 maths chapter 12 Exponents and Powers Chapter-13 CBSE NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions Chapter-14 NCERT solutions for class 8 maths chapter 14 Factorization Chapter-15 Solutions of NCERT for class 8 maths chapter 15 Introduction to Graphs Chapter-16 CBSE NCERT solutions for class 8 maths chapter 16 Playing with Numbers

## Importance of NCERT solutions for class 8 maths chapter 8 comparing quantities:

• To familiarise with the concept studied in the chapter NCERT solutions for class 8 maths chapter 8 comparing quantities are helpful.
• Solving homework exercise becomes easy with the solutions of NCERT class 8 maths chapter 8 comparing quantities in hand.
• Practicing NCERT solutions for class 8 maths chapter 8 comparing quantities helps to score well in the exam.